Question

Two solid bodies rotate about stationary mutually perpendicular intersecting axes with constant angular velocities ${\omega }_1=3rad/s$ and ${\omega }_2=4rad/s$. Find the angular velocity and angular acceleration of one body relative to the other.

Answer 1

The angular velocity is a vector as infinitesimal rotation commute. The the relative angular velocity of the body 1 with respect to the body 2 is clearly,

$\overrightarrow{{\omega }_{12}}=\overrightarrow{{\omega }_1}-\overrightarrow{{\omega }_2}$

as for relative linear velocity. The relative acceleration of 1 w.r.t. 2 is

${\left(\frac{d\overrightarrow{{\omega }_1}}{dt}\right)}_{S^{\prime }}$

where $S^{\prime }$ is a frame corotating with the second body and $S$ is a space fixed frame with origin coinciding with the point of intersection of the two axes,

but ${\left(\frac{d\overrightarrow{{\omega }_1}}{dt}\right)}_S={\left(\frac{d\overrightarrow{{\omega }_1}}{dt}\right)}_{S^{\prime }}+\overrightarrow{{\omega }_2}\times \overrightarrow{{\omega }_1}$

Since $S^{\prime }$ rotates with angular velocity $\overrightarrow{{\omega }_2}$. However ${\left(\frac{d\overrightarrow{{\omega }_1}}{dt}\right)}_S=0$ as the first body rotates with constant angular velocity in space, thus

$\overrightarrow{{\beta }_{12}}=\overrightarrow{{\omega }_1}\times \overrightarrow{{\omega }_2}$.

Note that for any vector $\overrightarrow{b}$, the relation in space forced frame $\left(k\right)$ and a frame $\left(k^{\prime }\right)$ rotating with angular velocity $\overrightarrow{\omega }$ is

$\frac{d\overrightarrow{b}}{dt}{|}_K=\frac{d\overrightarrow{b}}{dt}{|}_{K^{\prime }}+\overrightarrow{\omega }\times \overrightarrow{b}$